I'm trying to implicitly differentiate the following equation: $\frac{x+f(x)}{c+f(x)}=4$ where $c$ is a constant.
Here's what I get: $$ \frac{\mathrm d}{dx}\left(\frac{x+f(x)}{c+f(x)}\right)=\frac{\mathrm d}{\mathrm dx}(4) $$ $$ \frac{(1+f'(x))(c+f(x))-(x+f(x))(f'(x))}{(c+f(x))^2}=0 $$ $$ \frac{c+f(x)+cf'(x)+f(x)f'(x)-xf'(x)-f(x)f'(x)}{(c+f(x))^2}=0 $$ $$ c+f(x)+cf'(x)-xf'(x)=0 $$ $$ c+f(x)=xf'(x)-cf'(x) $$ $$ f'(x)=\frac{c+f(x)}{x-c} $$
However, when I isolate $f(x)$ first... $$ \frac{x+f(x)}{c+f(x)}=4 $$ $$ x+f(x)=4c+4f(x) $$ $$ x-4c=3f(x) $$ $$ \frac{x}{3}-\frac{4}{3}c=f(x) $$
... I get a different answer: $$ f'(x)=\frac{\mathrm d}{\mathrm dx}\left(\frac{x}{3}-\frac{4}{3}c\right) $$ $$ f'(x)=\frac{1}{3} $$
Hopefully someone can shed some light on where I'm going wrong.
Thanks!
From $\frac{x+f(x)}{c+f(x)}=4 $ find $f(x)$
$$4c+4f(x)=x+f(x)\\3f(x)=x-4c\\f(x)= \frac 13 (x-4c) $$ now you can easily see $f'(x)=\frac 13$ or put $f(x)$ into $f'(x)=\frac{c+f(x)}{x-c}$
$$f'(x)=\frac{c+\frac 13 (x-4c)}{x-c}=\frac 13 $$